C# Programming - Structures - Discussion
Discussion Forum : Structures - General Questions (Q.No. 17)
17.
Which of the following will be the correct output for the program given below?
namespace IndiabixConsoleApplication
{
struct Sample
{
public int i;
}
class MyProgram
{
static void Main(string[] args)
{
Sample x = new Sample();
Sample y;
x.i = 9;
y = x;
y.i = 5;
Console.WriteLine(x.i + " " + y.i);
}
}
}Discussion:
3 comments Page 1 of 1.
Bhupi said:
1 decade ago
Though the object is copied as it is but in the next line for the second object the value is changed so it will be a change in value for the object y's variable i.
Sark said:
1 decade ago
Because struct is a value type, when a instance assigned to another, all the elements will be copied to another location in stack, but vice versa when using class object.
See the example:
public class Sample
{
public int i;
}
public class MyProgram
{
public static void Main(string[] args)
{
Sample x = new Sample();
Sample y;
x.i = 9;
y = x;
y.i = 5;
Console.WriteLine(x.i + " " + y.i);
}
}
Result.
5 5.
See the example:
public class Sample
{
public int i;
}
public class MyProgram
{
public static void Main(string[] args)
{
Sample x = new Sample();
Sample y;
x.i = 9;
y = x;
y.i = 5;
Console.WriteLine(x.i + " " + y.i);
}
}
Result.
5 5.
Oli said:
7 years ago
@Saks.
The answer is right. It is a struct and passed as a copy. not a reference type.
See sample code
struct Sample
{
int i;
};
int _tmain(int argc, _TCHAR* argv[])
{
struct Sample x;
struct Sample y;
x.i = 9;
y = x;
y.i = 5;
cout << x.i + " " + y.i << endl;
return 0;
}
The answer is right. It is a struct and passed as a copy. not a reference type.
See sample code
struct Sample
{
int i;
};
int _tmain(int argc, _TCHAR* argv[])
{
struct Sample x;
struct Sample y;
x.i = 9;
y = x;
y.i = 5;
cout << x.i + " " + y.i << endl;
return 0;
}
(1)
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